The Unapologetic Mathematician

Mathematics for the interested outsider

The Metric Space of a Measure Ring

Let (\mathcal{S},\mu) be a measure ring. We’ve seen how we can get a metric space from a measure, and the same is true here. In fact, since we’ve required that \mu be positive — that \mu(E)=0 only if E=\emptyset — we don’t need to worry about negligible elements.

And so we write \mathfrak{S}=\mathfrak{S}(\mu) for the metric space consisting of the elements E\in\mathcal{S} with \mu(E)<\infty. This has a distance function \rho defined by \rho(E,F)=\mu(E\Delta F). We also write \mathfrak{S}=\mathfrak{S}(\mu) for the metric associated with the measure algebra associated with the measure space (X,\mathcal{S},\mu). We say that a measure space or a measure algebra is “separable” if the associated metric space is separable.

Now, if we set

\displaystyle\begin{aligned}f(E,F)&=E\cup F\\g(E,F)&=E\cap F\end{aligned}

then f:\mathfrak{S}\times\mathfrak{S}\to\mathfrak{S}, g:\mathfrak{S}\times\mathfrak{S}\to\mathfrak{S}, and \mu:\mathfrak{S}\to\mathbb{R} itself are all uniformly continuous.

Indeed, if we take two pairs of sets E_1, E_2, F_1, and F_2, we calculate

\displaystyle\begin{aligned}\mu\left((E_1\cup F_1)\setminus(E_2\cup F_2)\right)&=\mu\left((E_1\cup F_1)\cap(E_2^c\cap F_2^c)\right)\\&=\mu\left((E_1\cup F_1)\cap(E_2^c\cap F_2^c)\right)\\&=\mu\left((E_1\cap E_2^c\cap F_2^c)\cup(F_1\cap E_2^c\cap F_2^c)\right)\\&\leq\mu(E_1\cap E_2^c\cap F_2^c)+\mu(F_1\cap E_2^c\cap F_2^c)\\&\leq\mu(E_1\setminus E_2)+\mu(F_1\setminus F_2)\end{aligned}

Similarly, we find that \mu\left((E_2\cup F_2)\setminus(E_1\cup F_1)\right)\leq\mu(E_2\setminus E_1)+\mu(F_2\setminus F_1). And thus

\displaystyle\begin{aligned}\rho\left((E_1\cup F_1),(E_2\cup F_2)\right)&=\mu\left((E_1\cup F_1)\Delta(E_2\cup F_2)\right)\\&=\mu\left((E_1\cup F_1)\setminus(E_2\cup F_2)\right)+\mu\left((E_2\cup F_2)\setminus(E_1\cup F_1)\right)\\&\leq\mu(E_1\setminus E_2)+\mu(F_1\setminus F_2)+\mu(E_2\setminus E_1)+\mu(F_2\setminus F_1)\\&=\mu(E_1\Delta E_2)+\mu(F_1\Delta F_2)\\&=\rho(E_1,E_2)+\rho(F_1,F_2)\end{aligned}

And so if we have control over the distance between E_1 and E_2, and the distance between F_1 and F_2, then we have control over the distance between E_1\cup F_1 and E_2\cup F_2. The bounds we need on the inputs uniform, and so f is uniformly continuous. The proof for g proceeds similarly.

To see that \mu is uniformly continuous, we calculate

\displaystyle\begin{aligned}\left\lvert\mu(E)-\mu(F)\right\rvert&=\left\lvert\left(\mu(E)-\mu(E\cap F)\right)-\left(\mu(F)-\mu(F\cap E)\right)\right\rvert\\&=\left\lvert\mu\left(E\setminus(E\cap F)\right)-\mu\left(F\setminus(F\cap E)\right)\right\rvert\\&=\left\lvert\mu(E\setminus F)-\mu(F\setminus E)\right\rvert\\&\leq\mu(E\setminus F)+\mu(F\setminus E)\\&=\mu(E\Delta F)\\&=\rho(E,F)\end{aligned}

Now if (X,\mathcal{S},\mu) is a \sigma-finite measure space so that the \sigma-ring \mathcal{S} has a countable set of generators, then \mathfrak{S}(\mu) is separable. Indeed, if \{E_n\} is a countable sequence of sets that generate \mathcal{S}, then we may assume (by \sigma-finiteness) that \mu(E_n)<\infty for all n. The ring generated by the \{E_n\} is itself countable, and so we may assume that \{E_n\} is itself a ring. But then we know that for every E\in\mathfrak{S}(\mu) and for every positive \epsilon we can find some ring element E_n so that \rho(E,E_n)=\mu(E\Delta E_n)<\epsilon. Thus \{E_n\} is a countable dense set in \mathfrak{S}(\mu), which is thus separable.

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August 6, 2010 - Posted by | Analysis, Measure Theory


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